Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> ans;
void mswap(vector<int>& A, int i, int j) {
  swap(A[i], A[j]);
  ans.push_back({i, j});
  ans.push_back({j, i});
  ans.push_back({i, j});
}
void mxor(vector<int>& A, int i, int j) {
  A[i] ^= A[j];
  ans.push_back({i, j});
}
void gauss(vector<int>& A) {
  int N = A.size();
  for (int byte = 30, posx = 0; byte >= 0 && posx < N; --byte) {
    int pos;
    for (pos = posx; pos < N && !(A[pos] & (1 << byte)); ++pos)
      ;
    if (pos == N) continue;
    if (pos > posx) mswap(A, posx, pos);
    for (int i = posx + 1; i < N; ++i)
      if (A[i] & (1 << byte)) mxor(A, i, posx);
    ++posx;
  }
}
int main() {
  int N;
  scanf("%d", &N);
  vector<int> A(N), B(N);
  for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
  for (int i = 0; i < N; ++i) scanf("%d", &B[i]);
  gauss(B);
  vector<pair<int, int>> bops = ans;
  reverse(bops.begin(), bops.end());
  ans.clear();
  gauss(A);
  for (int i = 0; i < N; ++i) {
    for (int byte = 30; byte >= 0; --byte) {
      if ((A[i] ^ B[i]) & (1 << byte))
        for (int j = i; j < N; ++j) {
          if ((A[j] & (1 << byte)) && (A[j] < (1LL << (byte + 1)))) {
            mxor(A, i, j);
            break;
          }
        }
      if ((A[i] ^ B[i]) & (1 << byte)) {
        printf("-1\n");
        return 0;
      }
    }
  }
  ans.insert(ans.end(), bops.begin(), bops.end());
  bops.clear();
  printf("%d\n", int(ans.size()));
  for (auto p : ans) printf("%d %d\n", p.first + 1, p.second + 1);
}